A ball is thrown...

Question

A ball is thrown straight up, from 11 meters above the ground, with a velocity of 7 m/s. The equation below models the height of the ball:

h=−4t2+7t+11

a. Use the discriminant to determine the number and type of solutions for the quadratic equation.

b. Graph the equation to verify the result that the discriminant indicated.

c. Solve the equation by graphing, factoring, and the quadratic formula.

d. What do you notice about each method?

e. Describe what the solutions mean in context of the problem.

f. Describe the domain of this problem.

Answer 1

2 real solutions

the discriminant = 49+176= 225>0

if greater than zero, there's two real solutions

It's a downward opening parabola

h(t) =-4t^2 +7t + 11

y intercept = 11

x intercepts: -1 and 1 1/4

axis of symmetry is t=7/8

vertex is (7/8, h(7/8)) = (7/8, 14 1/16)= maximum point

(-4t+11)(t+1) =0

set each factor equal to zero and solve for t

-4t+11=0

t+1=0

or use the quadratic formula

t = 7/8 + or - (1/8)sqr225

= (7+ or- 15)/8 =-8/8 or 22/8

= -1 or 11/4

=-1 second or 2 3/4 seconds

it hits the ground in 2 3/4 seconds

it reaches a maximum in 7/8 of a second

maximum height = h(7/8) = 3 1/16 meters

domain is t=0 to t= 2 3/4